"f" is a continuous function such that:  `int_2^6f(x)` = 7      ;  and  `int_0^6f(x)` = 22   ; then,  `int_0^3f(x) = ??` ``

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The previous response makes the incorrect assumption that `int_0^3 f(x) dx = int_3^6 f(x) dx` if f(x) is a continuous function.

For example f(x) = x is a continuous function but `int_0^3 f(x) dx = x^2/2|_0^3 = 9/2 - 0 = 9/2` but `int_3^6 f(x) dx = x^2/2|_3^6 = 36/2 - 9/2 = 27/2`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may split the evaluation of definite integral of a continuous function, over a closed interval, such that:

`int_0^6 f(x) dx = int_0^2 f(x) dx + int_2^6 f(x) dx`

Since the problem provides the values of `int_0^6 f(x) dx`  and `int_2^6 f(x) dx` , you need to substitute these values such that:

`22 = int_0^2 f(x) dx + 7 => int_0^2 f(x) dx = 22 - 7 = 15`

`int_0^3f(x) dx = int_0^2 f(x) dx + int_2^3 f(x) dx`

`int_2^6 f(x) dx = int_2^3 f(x) dx + int_3^6 f(x) dx`

Considering, with respect to the property of symmetry, `int_0^3f(x) dx = int_3^6 f(x) dx` , you should come up with the following notations, such that:

`int_0^3f(x) dx = int_3^6 f(x) dx = u`

`int_2^3 f(x) dx = v`

Changing the notations yields:

`{(u = 15 + v),(7 = v + u):} `` => {(u-v = 15),(u+v = 7):} => ` 2`u = 22 => u = 11`

Since `u` represents `int_0^3f(x) dx` , hence `int_0^3f(x) dx = 11` .

Hence, evaluating the definite integral `int_0^3f(x) dx` of the continuous function f(x), using the property of linearity of integral of continuous function, yields `int_0^3f(x) dx = 11` .

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